© WJEC CBAC Ltd.
GCE MARKING SCHEME
MATHEMATICS - M1-M3 & S1-S3 AS/Advanced
SUMMER 2013
© WJEC CBAC Ltd.
INTRODUCTION The marking schemes which follow were those used by WJEC for the Summer 2013 examination in GCE MATHEMATICS. They were finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conferences were held shortly after the papers were taken so that reference could be made to the full range of candidates' responses, with photocopied scripts forming the basis of discussion. The aim of the conferences was to ensure that the marking schemes were interpreted and applied in the same way by all examiners. It is hoped that this information will be of assistance to centres but it is recognised at the same time that, without the benefit of participation in the examiners' conferences, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about these marking schemes. Paper Page M1 1 M2 9 M3 17 S1 23 S2 26 S3 29
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1
M1
Q Solution Mark Notes
1(a)
B1 (0, 20) to (8, 20)
Or (18, 6) to (40, 6)
B1 (8, 20) to (18, 6)
B1 completely correct with
all units and labels.
1(b) Deceleration = gradient of graph M1 any correct method
D = 818
620
A1 ft graph +/-
D = 1.4 ms-2
A1 cao
OR
Use of v = u + at, v=6, u=20, t=10 M1
6 = 20 + 10a A1 allow –a
a = -1.4 ms-2
Magnitude of acceleration = 1.4 ms-2
A1 cao
1(c) Distance AB = Area under graph M1 used. Oe
= (8 20) + 0.5(20 + 6)10 + (22 6) B1 any correct area, ft graph
= 160 + 130 + 132
= 422 m A1 cao
v ms-1
t s O 8 18
6
20
40
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2
Q Solution Mark Notes
2(a)
N2L applied to lift and person M1 dim correct equation,
forces opposing
(M + 64)g – 7500 = (M+64) 0.425 A1 correct equation
M = 736 A1
2(b)
N2L applied to person M1 64g and R opposing
Dim correct equation
64g – R = 64a A1 correct equation
R = 64 9.8 – 64 0.425
R = 600 N A1
R
64g
a
7500 N
(M+64)g N
a
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3
Q Solution Mark Notes
3(a) v2 = u
2 + 2as, v=0, a=(±)9.8, s=18.225 M1 oe used
0 = u2 – 2 9.8 18.225 A1
u = 18.9 A1 convincing
3(b) Use of s = ut + 0.5at2, s=(±)2.8, a=(±)9.8,
u=18.9 M1 oe
-2.8 = 18.9t + 0.5 (-9.8)t2 A1
4.9t2 – 18.9t – 2.8 = 0
7t2 – 27t -4 = 0
(7t + 1)(t – 4) = 0 m1 correct method for
solving quad equ seen
t = 4s A1 cao
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4
Q Solution Mark Notes
4
4(a) N2L applied to B
5g – T = 5a M1 dim correct equation
5g and T opposing.
T = 5 9.8 – 5 1.61 A1
T = 40.95 N A1 cao
R = 9g = (88.2 N) B1 si
F = 9g = (88.2) B1 si
N2L applied to A M1 dim correct equation
T and F opposing
T – F = 9a A1
T – 88.2 = 9 1.61
= 0.3 A1 cao
4(b) limiting friction = 9g = 9 0.6g = 5.4g B1
Limiting friction > 5g
Particle will remain at rest R1 oe
T = 5g = 49 N B1
A
B
R
F
9g
a
a
T
T
5g
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5
Q Solution Mark Notes
5
5(a)(i) Resolve vertically M1 all forces, no extras
R + 84 = 12g A1
R = 33.6 A1 cao
5(a)(ii) Moments about C M1 equation, no extra force
oe
12g 0.2 = 84(x – 0.8) B1 any correct moment
A1 correct equation
84x = 12g 0.2 + 84 0.8
x = 1.08 A1 cao
5(b) When about to tilt about C, RD = 0 M1 si
Moments about C m1 equation, no extra force
Mg 0.8 = 12g 0.2
M = 3 A1
A B C D
12g
0.8
R 84 N
x
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6
Q Solution Mark Notes
6.
6(a) Conservation of momentum M1 equation required, only 1
sign error.
2u + 5 0 = 2 (-2) + 5 3 A1 correct equation
u = 5.5 A1
6(b) Restitution M1 only 1 sign error
3 – (-2) = -e(0 – 5.5) A1 ft u
e = 11
10=0.909 A1 cao
6(c) Impulse = change of momentum M1 for P or Q
I = 5(3 – 0)
I = 15 (Ns) A1 + required
6(d) v’ = ev M1 used
v’ = 0.25 3
v’ = 0.75 ms-1
A1 + required
P Q
u 0
-2 3
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7
Q Solution Mark Notes
7.(a) Resolve M1 attempted
X = 85 – 40 + 75 cos B1 any correct resolution
X = 85 – 40 + 75 0.8 A1 all correct accept cos36.9
X = 105
Resolve M1 attempted
Y = 60 – 75 sin
Y = 60 – 75 0.6 A1 all correct, accept sin 36.9
Y = 15
R = 22 15105 M1
R = 752 = 106.066 N A1 cao
= tan-1
105
15
M1 allow reciprocal
= 8.13 A1 cao
7(b) N2L applied to particle M1 dim correct equation
752 = 5a
a = 152 = 21.21 ms-2
A1 ft R if first 2 M’s gained.
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8
Q Solution Mark Notes
8. Area from AD from AB
APCD 48 3 4 B1
PBC 24 8 8/3 B1
Circle 4 3 3 B1
Lamina (72-4) x y B1 areas
8(a) Moments about AD M1 equation
48 3 + 24 8 = 4 3 + (72 - 4)x A1 ft table
x = 5.02 cm A1 cao
Moments about AB M1 equation
48 4 + 24 8/3 = 4 3 + (72 - 4)y A1 ft table
y = 3.67 cm A1 cao
8(b) AQ = 3.67 cm B1 ft y
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9
M2
Q Solution Mark Notes
1(a) Loss in KE = 0.5mv2
= 0.5 8 72 M1 Corr use of KE formula
= 196J A1
1(b) Work energy principle M1 correct use
196 = F 15 A1 ft loss in KE
F = R
= 8g = (78.4) B1
Therefore 196 = 78.4 15
= 6
1 A1 ft loss in KE. Isw
OR
Use of v2=u
2+2as
0=72 + 2a×15 (M1)
a = -1.633
Use F = ma
-F = 8× -1.633 (M1)
F = 8g (B1)
= 6
1
g8
06713
(A1)p
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10
Q Solution Mark Notes
2(a) r = tv d M1 use of integration
r = ttt dj32i313 2
r = j2i32
13 32 tttt
+ ( c) A1 A1 one for each coefficient
When t = 0, m1 use of initial conditions
c = 2i + 7j A1 ft r
r = (6.5t2 – 3t + 2)i + (2t + t
3 + 7)j
2(b) a = t
v
d
d M1 use of differentiation
= 13i + 6tj A1
2(c) We require v.(i – 2j) = 0 M1 used
v.(i – 2j) = (13t – 3) – 2(2 + 3t2) M1 allow sign errors
= -6t2 + 13t – 7 A1 any form
6t2 – 13t + 7 = 0
(6t – 7)(t – 1) = 0 m1 method for quad equation
Depends on both M’s
t = 1, 7/6 A1
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11
Q Solution Mark Notes
3(a)(i) Initial horizontal speed = 15cos B1
= 15 0.8
= 12 ms-1
Time of flight = 9/12 M1
= 0.75s A1 any correct form
3(a)(ii) Initial vertical speed = 15 sin B1
= 15 0.6
= 9 ms-1
Use of s = ut + 0.5at2, u=9(c), a=(±)9.8,
t=0.75(c) M1
s = 9 0.75 – 0.5 9.8 0.752
s = 3.99375 m A1 si
Height of B above ground = 4.99375 m A1 ft s
3(b) use of v2 = u
2 + 2as, u=9, a=(±)9.8, s=-1 M1 allow sign errors
v2 = 9
2 + 2(-9.8)(-1) A1
v2 = 100.6
uH = 12 B1 ft candidate’s value
Speed = 6.100122 m1
Speed = 15.64 ms-1
A1 cao
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12
Q Solution Mark Notes
4(a) Resolve vertically M1 dim correct
Rsin = Mg A1
sin = 5
3 B1
R = Mg 3
5
R = 5Mg/3 A1 answer given, convincing.
4(b) N2L towards centre M1 dim correct
Rcos = Ma A1
3r
8gM
5
4
3
Mg5
CP = r = 2 A1
3
4
r
Height M1 use of similar triangles
Height = m3
8 A1 ft candidate’s r if first M1
given.
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13
Q Solution Mark Notes
5(a) 0 < t < 6 B1 B1
5(b) Distance t = 6 to t = 9 = 9
6
2 d122 ttt M1 use of integration
Limits not required
Distance = [ 2t3/3 – 6t
2]9
6 A1 correct integration
= 72
Distance t = 0 to t = 6 = - 6
0
2 d122 ttt
Distance = -[ 2t3/3 – 6t
2]
60
= -[-72]
= 72 A1 or for the other integral
Required distance = 72 + 72 m1
= 144 A1 cao
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14
Q Solution Mark Notes
6(a) T = P/v M1 used
T = 20
100060
T = 3000 N A1
6(b) Apply N2L to car and trailer M1 dim correct equation
All forces present
T – (1500+500)gsin -(170+30) = 2000a A2 -1 each error
3000 – 2000 9.8 14
1 -200 = 2000a
a = 0.7 ms-2
A1 convincing
6(c) N2L applied to trailer M1 dim correct, all forces
T – 500gsin -30 = 500a A2 -1 each error
T = 500 9.8 14
1 + 30 + 500 0.7
T = 730 N A1
OR
N2L applied to car (M1) dim correct, all forces
3000 - 1500gsin - 170 - T = 1500 0.7 (A2) -1 each error
T =3000-15009.814
1-170-1500 0.7
T = 730 N (A1)
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15
Q Solution Mark Notes
7(a) PE at start = - 2 9.8 0.7 M1 mgh used
A1 allow 0.7, (1.2+x), (0.5+x), 1.2,
0.5, x.
= -13.72 J
PE at end = -2 9.8 (1.2 + x)
= -23.52 – 19.6x
EE at end = 21
360
2
1
x
2 M1 use of formula
A1
EE at end = 150x2
Conservation of energy M1 equation, all energies
150x2 – 19.6x – 23.52 = -13.72 A1 correct equation any form
150x2 – 19.6x – 9.8 = 0
x = 0.33 A1 cao
7(b) KE at end = 0.5 2v2 B1
= v2
PE at end = -2 9.8 1.2
= -23.52
Conservation of energy M1 equation, no EE
v2 – 23.52 = -13.72 A1 correct equation, any form
v2 = 9.8
v = 3.13 ms-1
A1
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16
Q Solution Mark Notes
8(a) Conservation of energy M1 equation required
0.5mu2 + mgrcos = 0.5mv
2 + mgrcos A1 KE
A1 PE
0.5 3 52 + 3 9.8 4 0.8 =
0.5 3 v2 + 3 9.8 4 cos
75 + 188.16 = 3v2 + 235.2cos
v2 = 87.72 – 78.4cos
v = (87.72 – 78.4cos) A1 cao
8(b) N2L towards centre M1 dim correct, all forces
mgcos - R = ma A1
R = 3 9.8cos - 4
3(87.72 – 78.4cos) m1 substitute, v
2/r
R = 29.4cos - 65.79 + 58.8cos
R = 88.2cos - 65.79
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17
M3
Q Solution Mark Notes
1(a)(i) Apply N2L to particle M1 dim correct equation
ma = -mg – 3v A1
2t
v
d
d = -19.6 – v
1(a)(ii)
tv
vd
6.19
d2 M1 sep. of variables
2ln v6.19 = -t + (C) A1 correct integration
t = 0, v = 24.5 m1 use of initial conditions
C = 2ln 1.44 A1 ft no 2,1/2.
-t = 2ln1.44
6.19 v
e-t/2
= 1.44
6.19 v m1 inversion ln to e
v = 44.1 e-t/2
– 19.6 A1 cao
1(b) At maximum height, v = 0 M1 si
t = -2ln1.44
6.19
= 2 ln(2.25) = 1.62 s A1 ft similar expression
1(c) t
x
d
d = 44.1 e
-t/2 – 19.6 M1 v =
t
x
d
d used
x = -88.2 e-t/2
– 19.6t (+ C) A1 ft correct integration
When t = 0, x = 0 m1 use of initial conditions
C = 88.2
x = 88.2 – 88.2 e-t/2
– 19.6t A1 ft one slip
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18
Q Solution Mark Notes
2(a) Amplitude a = 0.5 B1
2(b) Period =
2 = 2 M1 si
= A1
Maximum acceleration = a2 = 0.5 2
B1 ft amplitude a.
Occurs at end points of motion B1
2(c) Let x = acos(t) M1
-0.25 = 0.5cos(t) m1
cos(t) = -0.5
t = 3
2
t = 3
2 A1 cao
2(d) v2 = 2
(a2 – x
2), x = 0.3, = M1
v2 = 2
(0.52 – 0.3
2) A1 ft
v2 = 2
0.42
v = (±)0.4
speed = 0.4 A1 cao
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19
Q Solution Mark Notes
3(a)(i) Apply N2L to P M1
2a = -8x – 10v A1
2
2
d
d
t
x = -4x - 5
t
x
d
d
3(a)(ii) 2
2
d
d
t
x + 5
t
x
d
d+ 4x = 0
Auxiliary equation m2 + 5m +4 = 0 B1
(m + 4)(m + 1) = 0
m = -4, -1 B1
CF x = Ae-t + Be
-4t B1 ft values of roots
When t = 0, x = 2, t
x
d
d= 3 M1 use of initial conditions
2 = A + B
t
x
d
d = -Ae
-t - 4 Be
-4t B1
3 = -A – 4B A1 both equations correct
Adding m1 solving simultaneously
5 = -3B
B = -3
5
A = 2 + -3
5 =
3
11
x = 3
11 e
-t -
3
5 e
-4t A1 cao
3(b) Try x = at + b M1
t
x
d
d = a
5a + 4(at + b) = 12t – 3 A1
4a = 12 m1 comparing coefficients
a = 3
5a + 4b = -3
15 + 4b = -3
4b = -18
b = -2
9
General solution x = Ae-t + Be
-4t + 3t -
2
9 A1 cao
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20
Q Solution Mark Notes
4 Initial speed of A just before impact = v
v2 = u
2 + 2as, u=0, a=(±)9.8, s =(1.8-0.2) M1
v2 = 0 + 2 9.8 1.6 A1
v = 5.6 ms-1
A1 cao
Impulse = Change in momentum M1 used
Applied to B
J = 3v B1
Applied to A
J = 5 5.6 – 5v A1 ft c’s answer in (a)
Solving
3v = 28 – 5v m1
8v = 28
v = 3.5 ms-1
A1 cao
J = 10.5 Ns A1 cao
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21
Q Solution Mark Notes
5(a) N2L applied to particle
0.25 a = 12
5
x M1
12
20
d
d
xx
vv M1 a = v
x
v
d
d
xx
vv d12
210d
M1 separating variables
2
1v
2 = 10 ln 12 x + C A1 correct integration ln
A1 LHS correct
When x = 0, v = 4 m1 use of boundary cond.
All 3 M’s awarded
8 = 10 ln(1) + C
C = 8
v2 = 20 ln 12 x + 16
ln 12 x = 20
1( v
2 – 16)
2x + 1 = 16050 2 ve m1 inversion, 3 M’s awarded
x = 0.5( 16050 2 ve – 1) A1 cao any equivalent exp.
5(b) v = 6
x = 0.5(e0.05(36 – 16)
– 1) M1 exp. with v2 needed
x = 0.5(e – 1)
x = 0.86 m A1 cao
5(c) a = 5
12
20
x = 5 M1
20 = 10x + 5
x = 1.5 A1
v2 = 20 ln(3 + 1) + 16 m1 substitution in expression
with v2.
= 20 ln4 + 16
v = 6.61 ms-1
A1 cao
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22
Q Solution Mark Notes
6
6(a) Moments about A M1 equation, no extra forces
No missing forces
6g 2 + 3g 3 = T 4sin A2 -1 each error
4 5
3T = 21g
T = 4
35g = 85.75 N A1 cao
6(b) Resolve vertically M1 equation, all forces, no
extra force
Tsin + Y = 9g A1
Y = 9g - 4
35g
5
3
Y = 4
15g = 36.75 N A1 cao
Resolve horizontally
Tcos = X M1 equation, all forces,
No extra force
X = 4
35g
5
4
X = 7g = 68.6 N A1 cao
6(b)(i) Magnitude of reaction at wall
= 22 7536668 M1
= 77.82 N A1 ft X and Y
6(b)(ii) = X
Y M1 used
= 74
15
=
28
15 A1 ft X and Y if answer<1.
A B
D
C X
Y
3
2 1 1
T
6g 3g
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23
S1
Ques Solution Mark Notes
1(a)
(b)
P(AB) = P(A) + P(B)
P(B) = 0.4 – 0.25 = 0.15
P(AB) = P(A) + P(B) – P(A)P(B)
0.4 = 0.25 + P(B) – 0.25P(B)
P(B) = 0.15/0.75 = 0.2
M1
A1
M1
A1
A1
Award M1 for using formula
Award M1 for using formula
2(a)
(b)
(c)
P(1 of each) =
3
10
1
2
1
3
1
5or 6
8
2
9
3
10
5
= 4
1
P(3 war) =
3
10
3
5or
8
3
9
4
10
5
= 12
1
P(3 cowboy) =
3
10
3
3or
8
1
9
2
10
3
= 120
1
P(3 the same) = 120
11
120
1
12
1
M1A1
A1
M1
A1
B1
M1A1
M1A0A0 if 6 omitted
Special case : if they use an
incorrect total, eg 9 or 11, FT
their incorrect total but subtract
2 marks at the end
FT previous values
3
20)( XE
Var(X) = 4 (SD = 2)
)(YE 20a + b = 65
Var(Y) = 4a2 = 36
a = 3
b = 5
B1
B1
B1
B1
B1
B1
Accept SD(Y) = 2a = 6
Must be justified by solving the
two equations
4(a)(i)
(ii)
(iii)
(b)(i)
(ii)
B(20,0.25)
P(3 ≤ X ≤ 9) = 0.9087 – 0.0139 or 0.9861 – 0.0913
= 0.8948
P(X = 6) = 146 75.025.06
20
= 0.169
Let Y denote the number of throws giving ‘8’
Then Y is B(160,0.0625) Poi(10).
P(Y = 12) = !12
10e
1210
= 0.0948
P(6≤Y≤14) = 0.9165 – 0.0671 or 0.9329 – 0.0835
= 0.8494 cao
B1
B1B1
B1
M1
A1
B1
M1
A1
B1B1
B1
B must be mentioned and the
parameters n and p must be seen
or implied somewhere in the
question
FT an incorrect p except for the
last three marks
M0 if no working seen
M0 if no working seen
Accept the use of tables
Correct values only (no FT)
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24
5(a)
(b)
P(1) = 2
1
3
1
3
1
3
1
4
1
3
1
= 36
13 (0.361)
P(A|1) = 36/13
12/1
= 13
3 cao (0.231)
M1A1
A1
B1B1
B1
M1 Use of Law of Total Prob
(Accept tree diagram)
FT denominator from (a)
B1 num, B1 denom
6(a)
(b)
The sequence is MMMH si
Prob = 0.30.30.30.7 = 0.0189
The sequence is MHH or HMH si
Prob = 0.30.70.7 + 0.70.30.7 = 0.294
B1
M1A1 B1
M1A1
Award B1 for 0.147
7(a)
(b)
(c)(i)
(ii)
8
1
4
1
2
11kpx = 1
15
81
8
1248
kk
815
14
15
22
15
41
15
8)( XE
= 15
32 (2.13)
E( )8( 6415
116
15
24
15
41
15
8)2 X
Var(X) = 45.315
328
2
(776/225)
The possibilities are (1,1); (2,2); (4,4); (8,8) si
P 2222
2115
1
15
2
15
4
15
8
XX
= (0.378) 45
17
It follows that P 45
2821 XX
And therefore by symmetry P 45
1421 XX
M1
A1
M1
A1
M1A1
A1
B1
M1
A1
M1
A1
.
Convincing
Accept 3.46
FT their answer from (c)(i)
Do not accept any other method.
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25
8(a)
(b)
Let X denote the number of calls between 9am and
10 am so that X is Po(5)
P(X = 7) = !7
5e 75
= 0.104
We require
P(calls betw 9 and 10=7|calls betw 9 and 11=10)
= 10) 11 and 9between calls(
10) 11 and 9 b c AND 7 10 and 9 b c(
P
P
= 10) 11 and 9between calls(
3) 11 and 10 b P(c 7) 10 and 9 b c(
P
P
= !10
10e
!3
5e
!7
5e 10103575
(denom = 0.125)
= 0.117
B1
M1
A1
M1
A1
A1A1
A1
M0 no working
A1 numerator, A1 denominator
The denominator A1 can be
awarded if the M1 is awarded
9(a)
(b)
(c)(i)
(ii)
1d4
1
2
0
2
x
xk
112
2
0
3
xxk
112
82
k
4
3k
E(X) = xx
x d)16
3
4
3(
2
0
2
=
2
0
42
64
3
8
3
xx
= 0.75
tt
xF
x
d)16
3
4
3()(
2
0
=
x
tt
0
3
164
3
= 164
3 3xx
)5.0()5.1()5.15.0( FFXP
= 0.547
M1
A1
A1
M1A1
A1
A1
M1
A1
A1
M1
A1
M1 for xxf d)( , limits not
required until next line
M1 for the integral of xf(x), A1
for completely correct although
limits may be left until 2nd
line.
M1 for xxf d)(
A1 for performing the
integration
A1 for dealing with the limits
FT their F(x)
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26
S2
Ques Solution Mark Notes
1(a)(i)
(ii)
(b)(i)
(ii)
25.02
105.10
z
P(X 10.5) = 0.5987
x = 282.1
x
= 12.564
E(X + 2Y ) = 34
Var(X + 2Y ) = Var(X) + 4Var(Y)
= 40
We require P(X + 2Y < 36)
z = 32.040
3436
Prob = 0.6255
Consider U = 21321 YYXXX
E(U) = 3 10 – 2 12 = 6
Var(U) = 3 4 + 2 9 = 30
We require P(U < 0)
z = 10.130
60
Prob = 0.136
M1A1
A1
M1
A1
B1
B1
M1A1
A1
B1
M1A1
m1A1
A1
M0 for 2or 22
M1A0 for – 0.25 if final answer
incorrect
M0 no working
M1 for 2.326, 1.645, 2.576
Accept 12.6
FT their mean and variance
M0 no working
Do not FT their mean and
variance
2(a)
(b)
50
9980x (= 199.6)
SE of 50
4X (= 0.5656…)
95% conf limits are
199.6 1.96 0.5656…
giving [198.5, 200.7] cao
Width of 95% CI = n
492.3 si
We require
14
92.3 n
n > 245.86…
Minimum n = 246
B1
B1
M1A1
A1
B1
M1
A1
A1
M1 correct form, A1 correct z.
M0 no working
FT their z from (a)
Award M1A0A0 for 1.96
instead of 3.92
FT from line above if n > 50
© WJEC CBAC Ltd.
27
3(a)
(b)
GBGB HH :;: 10
75.538
430;25.60
8
482 GB xx
SE of diff of means= 8
5.7
8
5.7 22
(3.75)
Test statistic (z) = 75.3
75.5325.60
= 1.73
Prob from tables = 0.0418
p-value = 0.0836 Insufficient evidence to conclude that there is a
difference in performance between boys and girls.
B1
B1B1
M1A1
m1A1
A1
A1
B1
B1
FT their z if M marks gained
FT on line above
FT their p-value
4(a)
(b)
(c)
4.0:;4.0: 10 pHpH
Let X = No. supporting politician so that
X is B(50,0.4) (under H0) si
p-value = P(X 25|X is B(50,0.4))
= 0.0978
Insufficient evidence to conclude that the support
is greater than 40%.
X is now B(400,0.4) (under H0) ≈ N(160,96)
p-value = P(X 181|X is N(160,96))
z = 96
1605.180
= 2.09
p-value = 0.0183
Strong evidence to conclude that the support is
greater than 40%.
B1
B1
M1
A1
B1
B1
M1
m1A1
A1
A1
B1
M0 for P(X = 25) or P(X > 25)
M0 normal or Poisson approx
FT on p-value
Award m1A0A1A1 for incorrect
or no continuity correction
181.5 z = 2.19 p = 0.01426
181 z = 2.14 p = 0.01618
FT on p-value
5(a)
(b)(i)
(ii)
H0: = 1.2 : H1: < 1.2
Let X = number of accidents in 60 days
Then X is Poi(72) (under H0) ≈ N(72,72) si
Sig level = P(X 58| H0)
z = 72
725.58
= – 1.59
Sig level = 0.0559
X is now Poi(48) which is approx N(48,48) si
P(wrong conclusion) = P(X 59| = 0.8)
z = 48
485.58
= 1.52
P(wrong conclusion) = 0.0643
B1
B1
M1
m1A1
A1
A1
B1
M1
m1A1
A1
A1
Must be
Award m1A0A1A1 for incorrect
or no continuity correction
57.5 z = –1.71 p = 0.0436
58 z = –1.65 p = 0.0495
Award m1A0A1A1 for incorrect
or no continuity correction
59.5 z = 1.66 p = 0.0485
59 z = 1.59 p = 0.0559
© WJEC CBAC Ltd.
28
6(a)(i)
(ii)
(b)(i)
(ii)
E(C) = 2E(R)
= 2 7 = 14 (43.98)
Var(C) = 24 Var(R)
= 3
4 2 (13.16)
P(C 45) = P(R 45/2)
= 68
)62/45(
= 0.581 2RA
P(A 150) = P /150R
= 68
/1508
= 0.545
EITHER
E(A) = rr d 2
18
6
2
= 863
6r
= 3
148 (155)
OR
E(A) = )(πE 2R = 2)(E)var( RR
=
27
3
1π
= 3
π148 (155)
M1
A1
M1
A1
M1
A1
A1
M1A1
A1
A1
M1
A1
A1
M1
A1
A1
Accept the use of integration,
M1 for a correct integral and A1
for the correct answer
© WJEC CBAC Ltd.
1
S3
Ques Solution Mark Notes
1
29.0ˆ p si
ESE = 300
71.029.0 (= 0.02619..) si
95% confidence limits are
..02619.096.129.0
giving [0.24,0.34]
B1
M1A1
m1A1
A1
m1 correct form, A1 1.96
2
The possibilities are
3 red, 1 blue for which |X – Y| = 2
Therefore,
P(X – Y = 2) = 47
7
8
1
9
2
10
3 OR
4
10
1
7
3
3
= 30
1
2 red, 2 blue for which |X – Y| = 0
P(X – Y = 0) = 67
6
8
7
9
2
10
3 OR
4
10
2
7
2
3
= 10
3
1 red, 3 blue for which |X – Y |= 2
P(X – Y = –2) = 47
5
8
6
9
7
10
3 OR
4
10
3
7
1
3
= 2
1
0 red, 4 blue for which |X – Y| = 4
P(X – Y = – 4) = 7
4
8
5
9
6
10
7 OR
6
1
4
10
4
7
The distribution of |X – Y| is therefore
|X – Y| 0 2 4
Prob 3/10 8/15 1/6
M1A1
A1
B1
B1
B1
M1A1
FT if found as 1 - probs
FT their probabilities
© WJEC CBAC Ltd.
2
3(a)
(b)
(c)
UE of = 34.3
43.106092 x
8
3.349
8
43.10609 of UE
22
= 2.6275
DF = 8 si
t-value = 1.86
90% confidence limits are
9
6275.286.13.34
giving [33.3,35.3] cao
EITHER
Width of interval = 2.39
6275.22 t
So t = 2.96
For a 99% confidence interval, t = 3.355
Since 2.96 < 3.355, the confidence level is less
than 99%
OR
For 99% confidence interval, t = 3.355
99% confidence limits are
9
6275.2355.33.34
giving [32.5,36.1]
The given confidence interval is narrower than
this therefore its confidence level is less than 99%
B1
B1
M1
A1
B1
B1
M1A1
A1
M1
A1 B1
A1
B1
M1
A1
A1
No working need be seen
M0 division by 9
Answer only no marks
Answer only no marks
4(a)
(b)
The 5% critical value = 2000 + 1.645 120
2554
= 2007.6
The 10% critical value = 2000 + 1.282 120
2554
= 2005.9
The required range is therefore
(2005.9,2007.6)
No because of the Central Limit Theorem
AND THEN EITHER
which ensures the normality of the sample mean
OR
which can be used because the sample is large
M1
A1
M1
A1
A1
B1
B1
M1A0 for –
M1A0 for –
© WJEC CBAC Ltd.
3
5(a)
(b)
BABA HH :;: 10
75.55;25.55 yx si
...2415.36059
3315
59
183345 22
xs
...8347.26059
3345
59
186651 22
ys
[Accept division by 60 giving 3.1875 and 2.7875]
SE = 60
..8347.2
60
..2415.3
= (0.3182.., 0.3155..) si
Test stat = ..3182.0
25.5575.55
= 1.57 (1.58)
p-value = 0.116 (0.114) cao
Insufficient evidence for believing that the mean
weights are unequal.
B1
B1
M1A1
A1
M1
A1
m1
A1
A1
B1
FT 1 error in the means
Answer only no marks
FT their p-value
6(a)
(b)
3170,1.118,5075,175 2 xyyxx
5.2177/1.1181753170 xyS
7007/1755075 2 xxS
b = 311.0700
5.217
a = 7
..311.01751.118 = 9.10
SE of a = 7007
50751.0 2
(0.1017…)
95% confidence limits for α are
...1017.096.110.9
giving [8.9,9.3]
B2
B1
B1
M1
A1
M1
A1
M1A1
m1A1
A1
Minus 1 each error
FT 1 error in sums
FT their value of a
M1 correct form, A1 1.96
© WJEC CBAC Ltd.
4
7(a)
(b)(i)
(ii)
(c)(i)
(ii)
E pn
np
n
XEp
)(ˆ
Therefore unbiased.
n
pp
n
pnp
n
Xp
)1()1()(VarˆSE
22
2
22 E
ˆEn
Xp
= 2
2)](E[)(Var
n
XX
= 2
22)1(
n
pnpnp
(n
ppp
)1(2 )
2p therefore not unbiased
)(E)(E)]1([E 2 XXXX
= nppnpnp 22)1(
= 2)1( pnn
It follows that
)1(
)1(
nn
XX
is an unbiased estimator for 2p .
EITHER
By reversing the interpretation of success and
failure, it follows that
)1(
)1)((
nn
XnXn
is an unbiased estimator for 2q .
OR
222 21)1( pppq
Therefore an unbiased estimator for 2q is
)1(
)1(21
nn
XX
n
X
Since pq = p(1 – p) = 2pp
It follows that an unbiased estimator for pq
= )1(
)1(
nn
XX
n
X
= )1(
)(
nn
XnX
M1
A1
M1
A1
M1
m1
A1
A1
M1
A1
A1
A1
M1
A1
M1
A1
M1
A1
A1
This line need not be seen
Accept q for 1 – p
This line need not be seen
This expression need not be
simplified
GCE MATHEMATICS M1-M3 and S1-S3 MS SUMMER 2013
© WJEC CBAC Ltd.
WJEC 245 Western Avenue Cardiff CF5 2YX Tel No 029 2026 5000 Fax 029 2057 5994 E-mail: [email protected] website: www.wjec.co.uk